If an object moves along a straight line, its motion can be represented by a velocity-time graph. The gradient of the line is equal to the acceleration of the object.
The table shows what each section of the graph represents:
|Section of graph||Gradient||Velocity||Acceleration|
|D||zero||stationary (at rest)||zero|
Describe the motion of the vehicle in the graph at the three stages of its journey.
Between 0 s and 4 s the vehicle is accelerating.
This means that:
Between 4 s and 7 s the vehicle has a constant velocity of 8 m/s.
Between 7 s and 10 s, the vehicle is decelerating so:
The displacement of an object can be calculated from the area under a velocity-time graph.
The area under the graph can be calculated by:
Calculate the total displacement of the object whose motion is represented by the velocity-time graph below.
The displacement can be found by calculating the total area of the shaded sections below the line.
Find the area of the triangle:
Find the area of the rectangle:
base × height
(10 - 4) × 8 = 48 m2
Add the areas together to find the total displacement:
(16 + 48) = 64 m